Jazmine Sweeney

2022-04-30

I'm supposed to solve this using Laplace Transformations.

Salvador Ayala

Beginner2022-05-01Added 10 answers

Let f(x) denote the integral, and assume temporarily that x>0. This makes no difference since f(x) is even by definition. Then its Laplace transform Lf(s) defines a continuous function on $s\in (0,\mathrm{\infty})$ (in fact, it defines an analytic function on R(s)>0). Thus we may assume further that $s\ne 1$ and then rely on the continuity argument. Then

$\mathcal{L}f\left(s\right)={\int}_{0}^{\mathrm{\infty}}f\left(x\right){e}^{-sx}dx$

$={\int}_{0}^{\mathrm{\infty}}{\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{cos}\left(xt\right)}{1+{t}^{2}}{e}^{-sx}dtdx$

$\stackrel{\cdot}{=}{\int}_{0}^{\mathrm{\infty}}{\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{cos}\left(xt\right)}{1+{t}^{2}}{e}^{-sx}dxdt$

$={\int}_{0}^{\mathrm{\infty}}\frac{1}{1+{t}^{2}}\frac{s}{{s}^{2}+{t}^{2}}dt$

$=\frac{s}{1-{s}^{2}}{\int}_{0}^{\mathrm{\infty}}(\frac{1}{{s}^{2}+{t}^{2}}-\frac{1}{1+{t}^{2}})dt$

$=\frac{s}{1-{s}^{2}}\frac{\pi}{2}(\frac{1}{s}-1)=\frac{\pi}{2}\frac{1}{1+s}$

Here, the change of order of integration (∗) is justified by the dominated convergence theorem. Though we proved this for$s\ne 1$ , it remains valid by continuity argument as mentioned above. Then by the uniqueness of the Laplace transform, we find that

$f\left(x\right)=\frac{\pi}{2}{e}^{-x}$

Here, the change of order of integration (∗) is justified by the dominated convergence theorem. Though we proved this for

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$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$